\( 1 \leftarrow \quad \begin{array}{l}\text { Assume that adults have IQ scores that are normally distributed with a mean of } \mu=105 \text { and a standard } \\ \text { deviation } \sigma=15 \text {. Find the probability that a randomly selected adult has an IQ between } 89 \text { and } 121 \text {. } \\ \text { Click to view page } 1 \text { of the table. Click to view page } 2 \text { of the table. } \\ \text { The probability that a randomly selected adult has an IQ between } 89 \text { and } 121 \text { is } \\ \text { (Type an integer or decimal rounded to four decimal places as needed.) }\end{array} \).

To find the probability that a randomly selected adult has an IQ between 89 and 121, we need to standardize these values using the z-score formula: \[ z = \frac{(X - \mu)}{\sigma} \] For \(X = 89\): \[ z_{89} = \frac{(89 - 105)}{15} = \frac{-16}{15} \approx -1.067 \] For \(X = 121\): \[ z_{121} = \frac{(121 - 105)}{15} = \frac{16}{15} \approx 1.067 \] Next, we will look up these z-scores in the standard normal distribution table (or use a calculator). For \(z \approx -1.067\), the cumulative probability is approximately 0.1423. For \(z \approx 1.067\), the cumulative probability is approximately 0.8577. Now, to find the probability that a randomly selected adult has an IQ between 89 and 121, we subtract the two cumulative probabilities: \[ P(89 < IQ < 121) = P(Z < 1.067) - P(Z < -1.067) \approx 0.8577 - 0.1423 = 0.7154 \] Therefore, the probability that a randomly selected adult has an IQ between 89 and 121 is approximately **0.7154**.